Pre-Calculus - Rational Functions

Find the zeros and asymptotes for

$y = \frac{2x^2 + 5 x - 3 }{2x^2 + 3x - 2 }$ .

Zero: ; Asymptote:

Zeros: ; Asymptote:

Zero: $x = \frac{1}{2}$ ; Asymptotes: $x = 2, -\frac{1}{2}$

Zeros: $x = 2, \frac{1}{2}$ ; Asymptotes: $x = 3, \frac{1}{2}$

Zero: ; Asymptotes: $x = -2, \frac{1}{2}$

Explanation

To find the information we're looking for, we should factor this equation:

$y = \frac{(2 x - 1 )(x + 3 )}{(2 x - 1 )(x+ 2 )}$

This means that it simplifies to $y = \frac{x + 3 }{ x + 2 }$ .

When the equation is in the form of a fraction, to find the zero of the function we need to set the numerator equal to zero and solve for the variable.

$\x+3=0\x=-3$

To find the asymptote of an equation with a fraction we need to set the denominator of the fraction equal to zero and solve for the variable.

$\x+2=0\x=-2$

Therefore our equation has a zero at -3 and an asymptote at -2.

Solve this equation and check your answer:

$1+\frac{n+6}{n+1}=\frac{4}{n-2}$

No solution

$n=\frac{1\pm\sqrt{73}}{2}$

$n=\frac{-7\pm\sqrt{193}}{4}$

$n=\frac{1\pm \sqrt{145}}{4}$

Explanation

To solve this, first, find the common denominator. It is (n+1)(n-2). Multiply the entire equation by this:

$(n+1)(n-2)\cdot 1+(n+1)(n-2)\frac{n+6}{n+1}=(n+1)(n-2)\frac{4}{n-2}$

Simplify to get:

Expand to get:

Move all terms to one side and combine to get:

Use the quadratic formula to get:

$n= \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = \frac{1\pm \sqrt{1^2 - 4\cdot 2\cdot -18}}{4} = \frac{1\pm \sqrt{145}}{4}$

Find the zeros and asymptotes for

$y = \frac{2x^2 + 5 x - 3 }{2x^2 + 3x - 2 }$ .

Zero: ; Asymptote:

Zeros: ; Asymptote:

Zero: $x = \frac{1}{2}$ ; Asymptotes: $x = 2, -\frac{1}{2}$

Zeros: $x = 2, \frac{1}{2}$ ; Asymptotes: $x = 3, \frac{1}{2}$

Zero: ; Asymptotes: $x = -2, \frac{1}{2}$

Explanation

To find the information we're looking for, we should factor this equation:

$y = \frac{(2 x - 1 )(x + 3 )}{(2 x - 1 )(x+ 2 )}$

This means that it simplifies to $y = \frac{x + 3 }{ x + 2 }$ .

When the equation is in the form of a fraction, to find the zero of the function we need to set the numerator equal to zero and solve for the variable.

$\x+3=0\x=-3$

To find the asymptote of an equation with a fraction we need to set the denominator of the fraction equal to zero and solve for the variable.

$\x+2=0\x=-2$

Therefore our equation has a zero at -3 and an asymptote at -2.

Solve this equation and check your answer:

$1+\frac{n+6}{n+1}=\frac{4}{n-2}$

No solution

$n=\frac{1\pm\sqrt{73}}{2}$

$n=\frac{-7\pm\sqrt{193}}{4}$

$n=\frac{1\pm \sqrt{145}}{4}$

Explanation

To solve this, first, find the common denominator. It is (n+1)(n-2). Multiply the entire equation by this:

$(n+1)(n-2)\cdot 1+(n+1)(n-2)\frac{n+6}{n+1}=(n+1)(n-2)\frac{4}{n-2}$

Simplify to get:

Expand to get:

Move all terms to one side and combine to get:

Use the quadratic formula to get:

$n= \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = \frac{1\pm \sqrt{1^2 - 4\cdot 2\cdot -18}}{4} = \frac{1\pm \sqrt{145}}{4}$

Find the slant and vertical asymptotes for the equation

$y = \frac{2x^2 - 5 x + 1 }{x + 3 }$ .

Vertical asymptote: ; Slant asymptote:

Explanation

To find the vertical asymptote, just set the denominator equal to 0:

To find the slant asymptote, divide the numerator by the denominator, but ignore any remainder. You can use long division or synthetic division.

$\x+3\ |\overline{2 x^ 2 - 5 x + 1 }$

$\ \underline{-(2x^2+3x) }$

$\underline{-(-8x-24)}$

The slant asymptote is

Solve $\frac{(x+1)(x-2)}{(x-3)^2(x+4)^3}<0$

-4 < x < -1 or x > 2

-1 < x < 2

x < -4 or -1 < x < 2

x < -4 or x > 2

Explanation

Draw a number line, and label each value that causes $\frac{(x+1)(x-2)}{(x-3)^2(x+4)^3}$ to be equal to zero. Then, draw a vertical line through each to separate it into "zones."

Screen shot 2020 05 28 at 10.00.29 pm

There are five "zones." We want to select a sample value within each zone and plug it into our expression to see if it yields a positive or negative result. You don't need to fully find each answer, only if it will be positive or negative. For example:

Plug in f(0): $\frac{(0+1)(0-2)}{(0-3)^2(0+4)^3} \rightarrow \frac{(+)(-)}{(-)(-)(+)(+)(+)}\rightarrow -$

Therefore, in the zone that includes 0, all values of x are negative, or less than zero, and therefore satisfy this equation.

Continue plugging in values in the other four zones to determine other possible solutions of the equation.

Plug in f(-5): $\frac{(-5+1)(-5-2)}{(-5-3)^2(-5+4)^3}\rightarrow \frac{(-)(-)}{(-)(-)(-)(-)(-)}\rightarrow -$

Plug in f(-2): $\frac{(-2+1)(-2-2)}{(-2-3)^2(-2+4)^3}\rightarrow \frac{(-)(-)}{(-)(-)(+)(+)(+)}\rightarrow +$

Plug in f(2.5): $\frac{(2.5+1)(2.5-2)}{(2.5-3)^2(2.5+4)^3}\rightarrow \frac{(+)(+)}{(-)(-)(+)(+)(+)}\rightarrow +$

Plug in f(4): $\frac{(4+1)(4-2)}{(4-3)^2(4+4)^3}\rightarrow \frac{(+)(+)}{(+)(+)(+)(+)(+)}\rightarrow +$

Therefore, the zones that include -5 and 0 are the ones that satisfy $\frac{(x+1)(x-2)}{(x-3)^2(x+4)^3}<0$ .

The solution to this equation is x < -4 or -1 < x < 2.

Find the slant and vertical asymptotes for the equation

$y = \frac{2x^2 - 5 x + 1 }{x + 3 }$ .

Vertical asymptote: ; Slant asymptote:

Explanation

To find the vertical asymptote, just set the denominator equal to 0:

To find the slant asymptote, divide the numerator by the denominator, but ignore any remainder. You can use long division or synthetic division.

$\x+3\ |\overline{2 x^ 2 - 5 x + 1 }$

$\ \underline{-(2x^2+3x) }$

$\underline{-(-8x-24)}$

The slant asymptote is

Find the slant asymptote for

$y = \frac{2x^2 - 13 x - 7 }{ x - 7 }$ .

This graph does not have a slant asymptote.

Explanation

By factoring the numerator, we see that this equation is equivalent to

$y = \frac{(2x+1)(x - 7 )}{x - 7}$ .

That means that we can simplify this equation to .

That means that isn't the slant asymptote, but the equation itself.

is definitely an asymptote, but a vertical asymptote, not a slant asymptote.